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There are $n$ random oberservations. Suppose $n$ is even number. We are interested in the ratio of each two. So there are $C_n^2=n(n-1)/2$ kinds of ratios. To study the property of the ratios we need independent assumption. So we decide to partition the $C_n^2$ combinations into $n-1$ sets and consider each separately. From set standpoint, by independence I mean each combination in a group has no elements in common. Additionaly, the $n$ elements in a group fully cover $1:n$, as shown in the example below.

is it always possible to partition the $C_n^2=n(n-1)/2$ combinations into $n-1$ groups satisfying above requirements? for example $n=6$, $C_6^2=15$

all combinations are

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

an independent partition containing $n-1=5$ groups is

  • 12 34 56
  • 13 25 46
  • 14 26 35
  • 15 24 36
  • 16 23 45

I can prove that given a solution for $n$, $2n$ can be solved.

In addition, I can provide an algorithm for $n=2^k$

But what about any even number? anyone have ideas?

Thank you!

user2792830
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    What is the definition of "independent groups"? – coffeemath Mar 01 '14 at 23:50
  • sorry I did not make it clear. I re-edit it. I consider the independence of the ratio of combinations in a group, so the definition of independent groups should be "combinations in the group has no elements in common" – user2792830 Mar 02 '14 at 18:53

1 Answers1

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What you're looking for are called "round robins" and exist for any even $n$. The idea is one has $n$ players (or baseball teams, etc.) and wants to set up a tournament of "rounds" where during the total set of rounds each player plays each other player exactly once, and during any one round each player is involved in a match against one of the other players. There are systematic ways to set these up for any even total number $n$ of players, some of which are discussed here.

coffeemath
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  • Thank you so much! That's exactly what I am looking for. It helps alot. But I don't have the reputation to vote up...... – user2792830 Mar 03 '14 at 00:00
  • @user2792830 I believe you don't need a certain reputation to "accept" the answer, i.e. click on the check-mark to make it turn green (and doing so gives you some reputation points). However you may want to hold off on doing that since someone else may submit a more complete answer you find more informative. – coffeemath Mar 03 '14 at 02:20